A sump pump (used to drain water from the basement of houses built below the wat

Question

A sump pump (used to drain water from the basement of houses built below the water table) is draining a basement at the rate of 0.55 L/s, with an output pressure of 3.15 times 10^5 N/*m^2. The water enters a hose with a 3 A cm inside diameter and rises 2 A m above the pump. What is its pressure at this point N/m^2? You may neglect frictional losses. The hose goes over the foundation wall, losing 0.45 m in height, and widens to 4.21 cm in diameter. What is the pressure now in N/m^2 You may neglect frictional losses.

Explanation / Answer

AV = 0.55L/s
Po = 3.15*10^5 N/m^2
V1 = 0.55*10^-3 / pi(0.034/2)^2
V2 = 0.55*10^-3 / pi(0.041/2)^2
P + 1000*(2.4-0.45)*9.8 + 0.5*1000*0.606^2 = P' + 0.5*1000*0.000001741304572225
P = 120293.6

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